Chapter 33 Algebra on First Subconstitutent

Wednesday, April 21, 1993

Lemma 33.1 Let $Y = (X, \{R_i\}_{0\leq i\leq D})$ be a commutative scheme. Fix a vertex $x\in X$, write $E^*_i\equiv E^*_i(x)$, $M^* \equiv M^*(x)$, $T\equiv T(x)$. Then the following hold.

$(i)$ $E^*_0MM^* = E^*_0M$

$(ii)$ $E^*_0T = E^*_0M$.

$(iii)$ $TE^*_0T = ME^*_0M$.

$(iv)$ $E^*_0E_0E^*_0 = |X|^{-1}E^*_0$.

$(v)$ $E^*_0E_0E^*_0 = |X|^{-1}E^*_0$.

$(vi)$ Lines $(i)$-$(iv)$ hold if we interchange $(E_0, E^*_0)$, $(M, M^*)$.

Moreover, $ME^*_0M = M^*E_0M^*$.

Proof.

$(i)$ $\supseteq$: $1\in M^*$ implies $M\subseteq MM^*$.

$\subseteq$: Pick $\alpha\in E^*_0MM^*$. Show $\alpha\in E^*_0M$. Since $A_0, A_1, \ldots, A_D$ span $M$, and since $E^*_0, E^*_1, \ldots, E^*_D$ span $M^*$, without loss of generality we may assume that \[\alpha = E^*_0A_iE^*_j\] for some $i,j\in \{0, \ldots, D\}$.

Without loss of generality we may assume taht $i = j$, else $\alpha = 0$ by Lemma 20.3.

\[(E^*_hA_iE^*_j \neq O \Leftrightarrow p^j_{hi}\neq 0.)\]

Now \[\alpha = E^*_0A_i\left(\sum_{h=0}^D E^*_h\right) = E^*_0A_i \in E^*_0M.\]

$(ii)$ $\supseteq$: This is clear.

$\subseteq$: $E^*_0T$ is the minimal right ideal of $T$ containing $E^*_0$.

So, we just have to show that $E^*_0M$ is a right ideal of $T$ containing $E^*_0$.

It clearly contains $E^*_0$ since $I\in M$, and is a right ideal of $T$ by $(i)$, and the fact that $T$ is generated by $M$ abd $M^*$.

$(iii)$ By the transpose of $(ii)$,

\[TE^*_0 = ME^*_0,\] so, \[TE^*_0T = (TE^*_0)(E^*_0T) = ME^*_0E^*_0M = ME^*_0M.\]

$(iv)$ We have

\[E^*_0E_0E^*_0 = \frac{1}{|X|}E^*_0\left(\sum_{h=0}^D A_h\right)E^*_0 = \frac{1}{|X|}E^*_0A_0E^*_0 = |X|^{-1}E^*_0.\]

$(v)$ The first part is clear by using Lemma 20.3 $(ii)$,

\[E_hA^*_iE_j\neq O \Leftrightarrow q^j_{hi} \neq 0,\] and Lemma 22.1 $(iii)$, \[q^j_{0i} = \delta_{ij}.\]

Also, \[ME^*_0M = TE^*_0T = TE^*_0E_0E^*_0T \subseteq TE_0T = M^*E_0M^*,\] and \[M^*E_0M^* \subseteq ME^*_0M\] by dual argument. So, \[M^*E_0M^* = ME^*_0M.\]

This proves the lemma.

Lemma 33.2 Let $\Gamma = (X, E)$ be distance regular of diameter $D\leq 3$, $Q$-polynomial with respect to $E_0, E_1, \ldots, E_D$. Pick a vertex $x\in X$, write $E^*_i\equiv E^*_i(x)$, $M^* \equiv M^*(x)$, $T\equiv T(x)$.

$(i)$ $E^*_1MM^* = E^*M + E^*_1E_0M^* + E^*_1E_1M^*$.

$(ii)$ $E_1M^*M = E_1M^* + E_1E^*_0M + E_1E^*_1M$.

Proof.

$(i)$ View $E^*_{-1}$, $E^*_{D+1}$ as $O$.

View $\theta^*_{-1}$, $\theta^*_{D+1}$ as indeterminates.

Let $\Delta$ denote $\mathrm{RHS}$ in $(i)$.

$\supseteq$: $I\in M^*$ implies $M\subseteq MM^*$.

$\subseteq$: Suppose not. Then there exists \[\begin{equation} \alpha\in E^*_1MM^*\setminus \Delta. \tag{33.1} \end{equation}\]

Since $A_0, A_1, \ldots, A_D$ span $M$, since $E^*_0, E^*_1, \ldots, E^*_D$ span $M^*$, without loss of generality we may assume that \[\alpha = E^*_1A_iE^*_j\] for some $i,j\in \{0, \ldots, D\}$.

Observe $|i-j|\leq 1$, else $\alpha =0$ by Lemma 20.3.

Without loss of generality, assume $i+j$ is minimal subject to the above constraints.

First assume \[\begin{equation} j = i+1. \tag{33.2} \end{equation}\] Observe \[\begin{align} & E^*_1A_iE^*_{i+1} + E^*_1A_iE^*_i + E^*_1A_iE^*_{i-1}\\ & \qquad = E^*_1A_i\left(\sum_{h=0}^D E^*_h\right)\\ & \qquad = E^*_1A_i\\ & \qquad \in \Delta. \tag{33.3} \end{align}\]

Also, observe \[E^*_1A_iE^*_i, \; E^*_1A_iE^*_{i-1}\in \Delta\] by the minimality of $i+j$, so \[\alpha = E^*_1A_iE^*_{i+1}\in \Delta\] by (33.3). Hence, (33.2) cannot occur.

Since $|i-j|\leq 1$, \[\begin{equation} i\in \{j, j+1\}. \tag{33.4} \end{equation}\]

Observe \[\begin{align} & E^*_1A_{j+1}E^*_{j} + E^*_1A_jE^*_j + E^*_1A_{j-1}E^*_{j}\\ & \qquad = E^*_1\left(\sum_{h=0}^D A_h\right)E^*_j\\ & \qquad = |X|E^*_1E_0E^*_j\\ & \qquad \in \Delta, \tag{33.5} \end{align}\] and \[\begin{align} & \theta^*_{j+1}E^*_1A_{j+1}E^*_{j} + \theta^*_j E^*_1A_jE^*_j + \theta^*_{j-1}E^*_1A_{j-1}E^*_{j}\\ & \qquad = E^*_1\left(\sum_{h=0}^D \theta^*_h A_h\right)E^*_j\\ & \qquad = |X|E^*_1E_1E^*_j\\ & \qquad \in \Delta. \tag{33.6} \end{align}\]

Since $E^*_1A_{j-1}E^*_j\in \Delta$ by the minimality of $i+j$, so \[E_1A_{j+1}E^*_j + E^*_1A_jE^*_j \in \Delta,\] \[\theta^*_{j+1}E^*_1A_{j+1}E^*_{j+1} + \theta^*_jE^*_1A_jE^*_j \in \Delta.\]

But, $\theta^*_0, \theta^*_1, \ldots, \theta^*_D$ are distinct by Lemma 22.2 $(iv)$, so \[E^*_1A_{j+1}E^*_j, \; E^*_1A_jE^*_j\in \Delta.\] But $\alpha$ is one of these two matrices, so \[\alpha \in \Delta.\] Hence, (33.4) cannot occur either, and we have a contradiction.

$(ii)$ Dual argument.

Lemma 33.3 With the above notation, set \[\tilde{J}:= E^*_1JE^*_1, \quad \tilde{A}:= E^*_1AE^*_1.\]

$(i)$ $\tilde{J}^2 = k\tilde{J}$. $(k = \text{valency of $\Gamma$})$

$(ii)$ $\tilde{J}\tilde{A} = \tilde{A}\tilde{J} = a_1\tilde{J}$. $(a_1 = p^{1}_{11} \text{ for } \Gamma)$

$(iii)$ $E_1^*E_0E^*_1 = |X|^{-1}\tilde{J}$.

$(iv)$ $E^*_1E_1E^*_1 = |X|^{-1}(E^*_1(\theta^*_0-\theta^*_2)+\tilde{A}(\theta^*_{1}-\theta^*_2)+\tilde{J}(\theta^*_2))$.

Proof.

$(i)$ The first subconstituent has $k$ vertices.

$(ii)$ The first subconstituent is regular of valency $a_1$.

$(iii)$ Since $E_0 = |X|^{-1}J$,

\[E^*_1E_0E^*_1 = |X|^{-1}\tilde{J}.\]

$(iv)$ We have

\[\begin{align} E^*_1E_1E^*_1 & = E^*_1\left(|X|^{-1}\sum_{h=0}^D\theta^*_h A_h\right)E^*_1\\ & = |X|^{-1}(\theta^*_0E^*_1A_0E^*_1 + \theta^*_1E^*_1A_1E^*_1 + \theta^*_2E^*_1A_2E^*_1)\\ & = |X|^{-1}(\theta^*_0E^*_1 + \theta^*_1\tilde{A} + \theta^*_2E^*_1A_2E^*_1). \tag{33.7} \end{align}\] Also, \[\begin{align} \tilde{J} & = E^*_1JE^*_1\\ & = E^*_1A_0E^*_1 + E^*_1A_1E^*_1 + E^*_1A_2E^*_1\\ & = E^*_1 + \tilde{A} + E^*_1A_2E^*_1. \tag{33.8} \end{align}\] Eliminating the $E^*_1A_2E^*_1$ term in (33.7) using equation (33.8), we get $(iv)$.

Lemma 33.4 With the above notation,

$(i)$ $E^*_1T = E^*_1E_0M^* + E^*_1M + E^*_1E_1M^* + E^*_1E_1E^*_1M + \cdots$.

$(ii)$ $E^*_1TE^*_1 = \mathrm{Span}(E^*_1E_0E^*_1, E^*_1, E^*_1E_1E^*_1, (E^*_1E_1E^*_1)^2, \ldots)$.

$(iii)$ $E^*_1TE^*_1 = \mathrm{Span}(\tilde{J}, E^*_1, \tilde{A}, \tilde{A}^2, \ldots )$.

$(iv)$ $E^*_1TE^*_1$ is symmetric (in particular, commutative).

Proof.

$(i)$ $\supseteq$: Clear.

$\subseteq$: $E^*_1T$ is the minimal right ideal of $\Gamma$ that contains $E^*_1$.

$\mathrm{RHS}$ contains $E^*_1$, so show $\mathrm{RHS}$ is a right ideal of $T$.

Show $\mathrm{RHS}$ is closed with respect to multiplication on right by $M$, $M^*$.

We have \[E^*_1E_0M^*(M) = E^*_1E_0M^*, \; E^*_1E_0M^*(M^*) = E^*_1E_0M^*\] by dual of Lemma 33.1 $(i)$.

By Lemma 33.2, \[\begin{align} & E^*_1E_1E^*_1\cdots E^*_1M(M^*)\\ & \qquad = E^*_1E_1E^*_1\cdots E_1(E^*_1MM^*)\\ & \qquad = E^*_1E_1E^*_1\cdots E_1(E^*_1M+E^*_1E_0M^* + E^*_1E_1M^*)\\ & \qquad \in \mathrm{RHS}, \end{align}\] because \[E^*_1E_1E^*_1\cdots E_1E^*_1E_0M^* \subseteq E^*_1TE_0T = E^*_1M^*E_0M^* = E^*_1E_0M^*.\]

By Lemma 33.2, \[\begin{align} & E^*_1E_1E^*_1\cdots E_1M^*(M)\\ & \qquad = E^*_1E_1E^*_1\cdots E^*_1(_1M^*M)\\ & \qquad = E^*_1E_1E^*_1\cdots E^*_1(E_1M^*+E_1E^*_0M^* + E_1E^*_1M)\\ & \qquad \in \mathrm{RHS}, \end{align}\] because by the last part of Lemma 33.1, \[E^*_1E_1E^*_1\cdots E^*_1E_1E^*_0M \subseteq E^*_1TE^*_0T = E^*_1ME^*_0M = E^*_1E_0M^*.\]

$(ii)$ Multiply $(i)$ on the right by $E^*_1$, we have

\[\begin{align} E^*_1TE^*_1 & = E^*_1E_0M^*E^*_1 + E^*_1ME^*_1 + E^*_1E_1M^*E^*_1\\ & \quad + \cdots + E^*_1E_1\cdots E_1M^*E^*_1 + E^*_1E_1\cdots E^*_1ME^*_1\\ & = \mathrm{Span}(E^*_1E_0E^*_1, E^*_1, E^*_1E_1E^*_1, (E^*_1E_1E^*_1)^2, \ldots). \end{align}\]

HS MEMO

Note that by Lemma 29.1, \[\begin{align} E^*_1ME^*_1 & = \mathrm{Span}(E^*_1A_0E^*_1, E^*_1A_1E^*_1, E^*_1A_2E^*_1)\\ & = \mathrm{Span}(E^*_1, E^*_1E_1E^*_1, E^*_1E_0E^*_1). \end{align}\] Moreover, \[E^*_1\cdots E^*_1E_0E^*_1 \subseteq E^*_1TE_0TE^*_1 = E^*_1M^*E_0M^*E^*_1 \in \mathrm{Span}(E^*_1E_0E^*_1).\]

$(iii)$ By $(ii)$, $E^*_1TE^*_1$ is generated by $\tilde{J} = |X|E^*_1E_0E^*_1$ and $E^*_1E_1E^*_1$.

By Lemma 33.3 $(iv)$, $E^*_1TE^*_1$ is generated by $\tilde{J}, \tilde{A}$.

But, $\mathrm{Span}{\tilde{J}}$ is a $2$-sided ideal by Lemma 33.3 $(i)$, $(ii)$.

Hence, we have $(iii)$.

$(iv)$ $\tilde{A}$, $\tilde{J}$ are symmetric commuting matrices, we have the claim.